Do I have a problem? Basic proportion problem.

1. Aug 11, 2012

Life-Like

Hello, this is an odd question but when I compared how I solved it compared to how the book solved it I kinda got worried. It was a very basic problem I been going back and touching up on my basics before I start pre-cal and trig in college.

The problem was: After completing 7/10 of his math homework assignment, Josh has 15 more questions complete. What is the total number of questions on his assignment? The answer was 50.

I got that by

$\frac{7}{10}$=$\frac{(x-15)}{x}$ Followed by cross multiplying and dividing.

However the book does it simply by doing the subtraction 10-7 = 3 then setting it up as

$\frac{3}{10}$=$\frac{15}{x}$ Then cross multiply and divide.

I know someone will say "It works so you can do what you are doing". But is there a possibility this kind of over complication( haha not hardly complicated!) Can hurt me later on? The way I performed it seemed much more intuitive... I have a highschool record of up to ap calc and ap chem and I'm worried if I pursue engineering in college I'll have a hard time solving simple things because of my use to complexity?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 11, 2012

azizlwl

(3/10)x=15
x=50

3. Aug 11, 2012

HallsofIvy

Staff Emeritus
Do you meant "15 more questions to complete"? What you wrote could be interpreted as meaning has completed 15 questions. If he has already completed 7/10 of his assignment, he still has 3/10 left. If he has 15 questions more to complete, letting "x" be the total number of problems, (3/10)x= 15 so x= (10/3)(15)= 150/3= 50.

Okay, that is, they are calculating 1- 7/10= 10/10- 7/10= (10- 7)/10= 3/10 as I did.

As long as you have linear problems, you can set them up as "proportions" and get the correct answer.

4. Aug 11, 2012

Life-Like

Thanks guys

Last edited: Aug 11, 2012